Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 11)
11.
One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
Answer: Option
Explanation:
Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in | x | minutes. |
3 |
![]() |
1 | + | 3 | = | 1 |
x | x | 36 |
![]() |
4 | = | 1 |
x | 36 |
x = 144 min.
Discussion:
67 comments Page 4 of 7.
Anurag said:
1 decade ago
@Senthil.
Lets start afresh,
If you think in terms of speed/rate of filling the tank of 'l' Litres capacity, then the speed of slower pipe will be x litre/min and the speed of faster pipe will be 3x litre/min.
But, if you think in terms of time, then the slower pipe will take x min(or 3x min) and the faster one will take x/3 min(or x min) to fill the tank.
Now, if you consider the second case i.e.,consider in terms of time, then you will only need to find out how much portion of the tank's capacity each pipe can fill in 36 min.
Since, in 1 min the slower pipe can fill 'l'/x litres of the tank's capacity(In x min 'l' litre), and
In 1 min the faster one can fill 'l'/(x/3) litres of the tank's capacity(In x/3 min 'l' litre).
Therefore, in 36 min slower pipe will fill 36*('l'/x) liters, And
In 36 min faster pipe will fill 36*('l'/(x/3)) liters.
Since, both the pipes are contributing in filling the tank of capacity 'l' Litres.
Therefore,
36*('l'/x) + 36*('l'/(x/3)) = 'l'.
('l'/x) + ('l'/(x/3)) = 'l'/36.
(1/x) + (1/(x/3)) = 1/36.
(1/x) + (3/x) = 1/36.
Hope, it will be clear now.
Lets start afresh,
If you think in terms of speed/rate of filling the tank of 'l' Litres capacity, then the speed of slower pipe will be x litre/min and the speed of faster pipe will be 3x litre/min.
But, if you think in terms of time, then the slower pipe will take x min(or 3x min) and the faster one will take x/3 min(or x min) to fill the tank.
Now, if you consider the second case i.e.,consider in terms of time, then you will only need to find out how much portion of the tank's capacity each pipe can fill in 36 min.
Since, in 1 min the slower pipe can fill 'l'/x litres of the tank's capacity(In x min 'l' litre), and
In 1 min the faster one can fill 'l'/(x/3) litres of the tank's capacity(In x/3 min 'l' litre).
Therefore, in 36 min slower pipe will fill 36*('l'/x) liters, And
In 36 min faster pipe will fill 36*('l'/(x/3)) liters.
Since, both the pipes are contributing in filling the tank of capacity 'l' Litres.
Therefore,
36*('l'/x) + 36*('l'/(x/3)) = 'l'.
('l'/x) + ('l'/(x/3)) = 'l'/36.
(1/x) + (1/(x/3)) = 1/36.
(1/x) + (3/x) = 1/36.
Hope, it will be clear now.
Senthil said:
1 decade ago
Why can't we take x+3x=36 and how 1/x+3/x is taken?
Senthil said:
1 decade ago
Why can't we take x+3x=36 and why we are taking 1/x+3/x=1/36 explain this solution in detail?
Vaibhaw said:
1 decade ago
I have another solution,
Suppose slower pipe is taking X min, then in that time faster pipe will fill 3 tanks.
Now both start filling tanks,then when slower pipe will finishing filling till that time faster also done with filling 3 tanks.
Total number of tanks filled is Slower= 1, and faster = 3, sum total is 4 tanks, NOw time taken for filling 1 tank is 36, Therefore time taken for filling 4 tanks is 36*4 = 144.
Therefore it indicates same time taken by slower pipe also to complete fill the tanks.
Suppose slower pipe is taking X min, then in that time faster pipe will fill 3 tanks.
Now both start filling tanks,then when slower pipe will finishing filling till that time faster also done with filling 3 tanks.
Total number of tanks filled is Slower= 1, and faster = 3, sum total is 4 tanks, NOw time taken for filling 1 tank is 36, Therefore time taken for filling 4 tanks is 36*4 = 144.
Therefore it indicates same time taken by slower pipe also to complete fill the tanks.
Ahammed riyas said:
1 decade ago
If we take x+x/3 instead of 1/x+3/x we cannot get answer because the method of solving will be more complicated.
Raj said:
1 decade ago
First pipe x.
Second pipe 3 times faster then it is 3x.
Together will fill it in 36 min.
1/x+1/3x=1/36.
Then we will get x=48 minutes and 3x= 144 minutes faster pipe is 48 minutes Which takes less time and slower pipe is 144 minutes which takes more time.
Second pipe 3 times faster then it is 3x.
Together will fill it in 36 min.
1/x+1/3x=1/36.
Then we will get x=48 minutes and 3x= 144 minutes faster pipe is 48 minutes Which takes less time and slower pipe is 144 minutes which takes more time.
Aditi said:
1 decade ago
If we take faster one as x,
Then d slower one will become 3x,
Thus the work done by them is equal to 1/36.
Therefore,
1/x + 1/(3x) = 1/36.
(3x + x)/(3x * x) = 1/36.
4/3x = 1/36.
x = 48.
Thus the slower one will become 144, thats the solution for taking x and 3x as the time.
Then d slower one will become 3x,
Thus the work done by them is equal to 1/36.
Therefore,
1/x + 1/(3x) = 1/36.
(3x + x)/(3x * x) = 1/36.
4/3x = 1/36.
x = 48.
Thus the slower one will become 144, thats the solution for taking x and 3x as the time.
Anil kumar said:
1 decade ago
If slower pipe is x. Than faster 3x. Take both 36 min. means x = 9 let 9+3 = 12 and 12 square = 144.
Ravikanth said:
1 decade ago
The best approach to these kinda questions are the logic of replacing the variables with the numbers but be careful in doing so because one should be aware of the numbers which are being replaced to the variables should be satisfying the condition of the variables.
So guys the easiest approach to the above problem is to:
P1 is three times faster than P2 right?
So I will write down a statement as follows,
P1=3P2.
So am gonna replace these with numbers as below,
P1=3. If p1=3 the above statement becomes 3=3P2.
By the cancellation we get the value of P2 as 1,
P2=1.
If a tank is to be filled by the two pipes together in 36 minutes then the values P1=3 becomes 3L/M P2 = 1L/M.
So my next statement is,
P1+P2 = 3+1.
P1+P2 = 4L/M.
If both the pipes are working together then the tank is being filled at a rate of 4L/M.
So the time taken by the to pipes to fill the tank is 36M.
Then we need to multiply the combined rate of pipes with the time to get the capacity of the tank.
4L/M*36M=144L which the capacity of the tank.
Now coming to the last part of the question.
"slower pipe alone will be able to fill the tank in:"
I already mentioned the slower pipe as P2 and the rate of P2=1L/M so the time taken by P2 to fill the tank of 144L capacity is (tank capacity/P2)=144L/1L/M=144M.
And finally sorry for the big explanation and guys please pardon me if there are any grammatical errors.
In brief,
P1 = faster pipe P2 = slower pipe.
P1=3P2.
If P1 = 3 =>3 = 3P2 =>P2 = 1.
Time = 36M.
P1+P2 = 1+3 =>P1+P2 = 4.
As the units of the pipes are liters/minute L/M
Find tank capacity,
(P1+P2)*Time=4*36=>Tank capacity = 144L.
144L capacity tank will be filled by slower pipe at its slower rate right?
rate of slower pipe P2?
P2 = 1L/M.
Time=Tank capacity/rate of pipe P2 => Time = 144/1 = 144.
Which is the required answer.
So guys the easiest approach to the above problem is to:
P1 is three times faster than P2 right?
So I will write down a statement as follows,
P1=3P2.
So am gonna replace these with numbers as below,
P1=3. If p1=3 the above statement becomes 3=3P2.
By the cancellation we get the value of P2 as 1,
P2=1.
If a tank is to be filled by the two pipes together in 36 minutes then the values P1=3 becomes 3L/M P2 = 1L/M.
So my next statement is,
P1+P2 = 3+1.
P1+P2 = 4L/M.
If both the pipes are working together then the tank is being filled at a rate of 4L/M.
So the time taken by the to pipes to fill the tank is 36M.
Then we need to multiply the combined rate of pipes with the time to get the capacity of the tank.
4L/M*36M=144L which the capacity of the tank.
Now coming to the last part of the question.
"slower pipe alone will be able to fill the tank in:"
I already mentioned the slower pipe as P2 and the rate of P2=1L/M so the time taken by P2 to fill the tank of 144L capacity is (tank capacity/P2)=144L/1L/M=144M.
And finally sorry for the big explanation and guys please pardon me if there are any grammatical errors.
In brief,
P1 = faster pipe P2 = slower pipe.
P1=3P2.
If P1 = 3 =>3 = 3P2 =>P2 = 1.
Time = 36M.
P1+P2 = 1+3 =>P1+P2 = 4.
As the units of the pipes are liters/minute L/M
Find tank capacity,
(P1+P2)*Time=4*36=>Tank capacity = 144L.
144L capacity tank will be filled by slower pipe at its slower rate right?
rate of slower pipe P2?
P2 = 1L/M.
Time=Tank capacity/rate of pipe P2 => Time = 144/1 = 144.
Which is the required answer.
Harshdeep said:
1 decade ago
Another simpler approach:
(slower) pipe A takes 'x' min to fill complete tank alone.
Then pipe A in 1 min fills (1/x) portion of tank alone.
(faster) pipe B takes (x/3) min to fill the complete tank alone.
then pipe B in 1 min fills (3/x) portion of tank alone.
When both pipes together fills the tank then in 1 min they fill (1/x + 3/x) portion of tank.
This means complete tank can be filled in (x + x/3) minutes.
(4x/3) = 36.... so x = 144 min . Answer.
(slower) pipe A takes 'x' min to fill complete tank alone.
Then pipe A in 1 min fills (1/x) portion of tank alone.
(faster) pipe B takes (x/3) min to fill the complete tank alone.
then pipe B in 1 min fills (3/x) portion of tank alone.
When both pipes together fills the tank then in 1 min they fill (1/x + 3/x) portion of tank.
This means complete tank can be filled in (x + x/3) minutes.
(4x/3) = 36.... so x = 144 min . Answer.
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